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\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
=\(x^2+2xy+y^2+x^2-2xy+y^2\)
\(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left[\left(x-y\right)+\left(x+y\right)\right]^2\)
= \(\left(x-y+x+y\right)^2\)
\(=2x^2\)
c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)
\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)
= \(\left(x-y+z-z+y\right)^2=x^2\)
Dùng hằng đẳng thức thứ 2:
A= [(x+y+z)-(x+y)]2=z2
Chúc bạn học tốt!
Áp dụng HĐT thứ 2: (A - B)2 = A2 - 2AB + B2, ta có:
(x + y + z)2 - 2(x + y + z)(x + y) + (x + y)2 = [(x + y + z) - (x + y)]2
= z2
giải
Áp dụng hàng đẳng thức đáng nhớ :
a ) \(5.\left(x+2\right)\left(x-2\right)-\left(3-4x\right)^2\)
\(=5\left(x^2-2^2\right)-\left(9-24x+16x^2\right)\)
\(=5x^2-20-9+24x-16x^2\)
\(=-11x^2+24x-29\)
b ) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2\left(x^2-y^2\right)+\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
\(=2x^2-2y^2+2x ^2+2y^2=4x^2\)
Chúc bạn học tốt !!!
Bài làm
a) 5 . ( x + 2 )( x - 2 ) - ( 3 - 4x )2
= 5 . ( x2 - 22 ) - [ 32 - 2.3.4x + ( 4x )2 ]
= 5x2 - 20 - ( 9 - 24x + 16x2 )
= 5x2 - 20 - 9 + 24x - 16x2
= ( 5x2 - 16x2 ) - ( 20 + 9 ) + 24x
= -11x2 - 29 + 24x
b) 2( x - y )( x + y ) + ( x + y )2 + ( x - y )2
= 2( x2 - y2 ) + x2 + 2xy + y2 + x2 - 2xy + y2
= 2x2 - 2y2 + x2 + 2xy + y2 + x2 - 2xy + y2
= ( 2x2 + x2 + x2 ) + ( -2y2 + y2 + y2 ) + ( 2xy - 2xy )
= 4x2
# Học tốt #
a)(x+y)3-3xy(x+y)
\(=\left(x+y\right)\left(x^2+xy+y^2\right)-3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+xy+y^2-3xy\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2-4ab\)
\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]-4ab\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)-4ab\)
\(=2b.2a-4ab\)
\(=4ab-4ab=0\)
Ta có:
x x - y - y y - x = x 2 - x y - y 2 - x y = x 2 - x y - y 2 + x y = x 2 - y 2
Chọn (B) x 2 - y 2